Animated Solution for Mathematics - Vector Algebra: Let the vectors $\vec{a}=(1+t)\hat{i}+(1-t)\hat{j}+\hat{k}$, $\vec{b}=(1-t)\hat{i}+(1+t)\hat{j}+2\hat{k}$ and $\vec{c}=t\hat{i}-t\hat{j}+\hat{k},t\in \mathbb{R}$ be such that for $\alpha ,\beta ,\gamma \in \mathbb{R},\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0}\Rightarrow \alpha =\beta =\gamma =0$. Then, the set of all values of $t$ is :

Welcome, student. Today, we are diving into the heart of vector algebra. We are given three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, and a condition that looks intimidating at first glance: $\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0}⟹\alpha =\beta =\gamma =0$.

Do not let the notation scare you. This is the mathematical definition of Linear Independence. It means these vectors are 'free'—none of them can be created by combining the others.

They span a full three-dimensional volume. If you were to walk along $\vec{a}$, then $\vec{b}$, and then $\vec{c}$, you would be exploring the full depth of space, never trapped in a single plane.

Measuring Volume

How do we translate this 'freedom' into a calculation? We use the Scalar Triple Product.

Geometrically, if we treat these three vectors as the edges of a parallelepiped, their scalar triple product gives us the volume of that shape. If the vectors are linearly independent, they are non-coplanar, meaning they form a 3D shape with a non-zero volume.

Therefore, the condition for linear independence is simply that the scalar triple product must not be zero: $[\vec{a} \ \vec{b} \ \vec{c}] eq 0$. This is our gateway to the solution.

Simplifying the Determinant

We construct our determinant using the components of our vectors as rows:

Now, we could expand this directly, but that is the path of the novice. We are aiming for elegance. Let us use column operations to simplify.

Notice the second column. If we apply $C_2\to C_1+C_2$, the $t$ terms in the first two rows will cancel out beautifully. The determinant becomes:

Look at that! We have created a zero in the third row. Let us go further. By applying $R_1\to R_1-R_2$, we create another zero in the second column:

Now, expanding along the second column is trivial. We take the element $2$, multiply it by the determinant of the remaining $2\times 2$ matrix, and set it to non-zero:

Evaluating this, we get $2\times (2t-(-t))=2\times (3t)=6t$. Thus, our condition is $6t eq 0$.

This implies $t eq 0$. Since $t$ is any real number, the set of all values for $t$ is simply $\mathbb{R}-\{0\}$. You have successfully navigated the geometry and the algebra. Well done.